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Regarding the same homework problem as this question:

A curve (projective, irreducible, smooth) $V$ is covered by two affine pieces (with respect to different embeddings) which are affine plane curves with equations $y^2 = f(x)$ and $v^2 = g(u)$ respectively, with $f$ a square-free polynomial of even degree $2n > 4$ and $u = 1/x$, $v = y/x^n$ in $k(V)$. Determine the polynomial $g(u)$ and show that the canonical class has degree $2n - 4$. Why can we not just say that $V$ is the zero locus of the homogenization of $y^2 = f(x)$?

Here $k$ is an algebraically closed field of characteristic $0$.

I'm having a hard time wrapping my head around the setting of this problem.

For one, could $V$ live in $\mathbb P^n$ for any $n$?

When the question says "affine pieces", one way to interpret this is to take $V \cap U_0, \dots V \cap U_n$, where $U_i$ are the standard affine patches of $\mathbb P^n$, and take $x$ and $y$ to be coordinates resulting from dehomogenization. However, since the question talks about embeddings, and indicates that $x,y,u,v$ are elements of $k(V)$, another way I think of interpreting this is that $x,y,u,v$ are general rational functions on $V$, and $V = U_1 \cup U_2$ so that e.g. $U_1 \to \mathbb P^2$ given by $(x:y:1)$ is an isomorphism (onto its image). Which way is intended?

From the given equations we get $y^2 = x^{2n}g(1/x)$ on the second affine piece. The natural next step is to declare that $f(x) = x^{2n}g(1/x)$, and then get $g$. But why should this hold?

ChrisOverflow
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1 Answers1

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The second method, taking $x,y,u,v$ as rational functions on $V$, so that $U_1\to \Bbb P^2$ by $[x:y:1]$ is an isomorphism on to its image and $U_2\to\Bbb P^2$ by $[u:v:1]$ is an isomorphism on to its image, is the intended one.

The reason that the equations $u=1/x$, $v=y/x^n$, $y^2=f(x)$, and $v^2=g(u)$ hold is because we supposed them to hold. We could work out the exact open sets where each of these is valid and hold, and worry about restricting to the right open sets when combining relations, but of course all of these are perfectly good equations inside the field of rational functions. So we can do arithmetic with them there and use those results, which makes things a little simpler for us since that's less baggage to carry around. Once we have a relation in the field of rational functions, this holds everywhere the involved functions are defined.

(By the way, I'd get $g(u)$ slightly differently - starting from $y^2=f(x)$, use $v=y/x^n$ to write $y=x^nv$ so that $x^{2n}v^2=f(x)$ and then $v^2=\frac{f(x)}{x^{2n}}$ and that RHS is a polynomial in $u$. That polynomial must be $g$ because otherwise we have that a polynomial in $u$ is identically zero on $V$, giving that $U_2$ is a point, contradicting the assumption that $V$ was one-dimensional.)

To answer the question "could $V$ live in $\Bbb P^n$ for any $n$?", let me first point out that the answer doesn't matter. It's nice of the author to give you the parenthetical that the curve is projective, but this is automatic for a curve, and one never really needs this fact to do everything they're asking of you. If $V$ was not a curve and instead a higher-dimensional variety, it may be the case that it is not projective at all, but we could still write it as a union of finitely many affine pieces, and we could still do something very much like this if given equations for certain affine charts.

To be (potentially annoyingly) specific about what $\Bbb P^n$ work for $V$ to live in, it's clear that it can't live in $\Bbb P^0$, some work with the canonical class will show that it isn't $\Bbb P^1$ so it can't live in $\Bbb P^1$, some more involved classification work will show that it can live in $\Bbb P^2$ sometimes, and some more theory of algebraic curves will show that it can always live in $\Bbb P^n$ for $n\geq 3$.

KReiser
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